[Leetcode] 742. Closest Leaf in a Binary Tree

最新推荐文章于 2020-09-18 02:52:31 发布

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本文介绍了一种高效算法，用于解决在给定二叉树中寻找目标节点最近叶子节点的问题。通过构建额外的哈希表来记录节点间的父子关系，并运用BFS策略，确保算法能准确找到距离目标节点最近的叶子节点。

题目：

Given a binary tree where every node has a unique value, and a target key k, find the value of the nearest leaf node to target k in the tree.

Here, nearest to a leaf means the least number of edges travelled on the binary tree to reach any leaf of the tree. Also, a node is called a leaf if it has no children.

In the following examples, the input tree is represented in flattened form row by row. The actual root tree given will be a TreeNode object.

Example 1:

Input:
root = [1, 3, 2], k = 1
Diagram of binary tree:
          1
         / \
        3   2

Output: 2 (or 3)

Explanation: Either 2 or 3 is the nearest leaf node to the target of 1.

Example 2:

Input:
root = [1], k = 1
Output: 1

Explanation: The nearest leaf node is the root node itself.

Example 3:

Input:
root = [1,2,3,4,null,null,null,5,null,6], k = 2
Diagram of binary tree:
             1
            / \
           2   3
          /
         4
        /
       5
      /
     6

Output: 3
Explanation: The leaf node with value 3 (and not the leaf node with value 6) is nearest to the node with value 2.

Note:

root represents a binary tree with at least 1 node and at most 1000 nodes.
Every node has a unique node.val in range [1, 1000].
There exists some node in the given binary tree for which node.val == k.

思路：

由于在题目中，要求“travel”不仅可以从父节点到子节点，而且还可以从子节点到父节点，所以实际上是把树作为一个图来处理了。那么我们怎么做呢？我们可以额外定义一个哈希表，建立从子节点到父节点的映射。然后就可以采用BFS方法，从值为k的节点开始遍历，找到第一个叶子结点，即为我们所求。

我开始实现的代码总是无法通过大数据测试，找不到原因。后来才发现，不同节点可能会被多次加入队列，造成了重复计算。所以我后来定义了另外一个哈希表，记录已经被访问过的节点。这样节点就不会被重复访问了，可以顺利通过所有测试数据。

代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int findClosestLeaf(TreeNode* root, int k) {
        TreeNode *node;
        unordered_map<TreeNode*, TreeNode*> parent;
        constructParent(root, parent, k, node);
        queue<pair<TreeNode*, int>> q;
        q.push(make_pair(node, 0));
        if (!node->left && !node->right) {
            return node->val;
        }
        unordered_set<TreeNode*> visited;
        visited.insert(node);
        while (!q.empty()) {
            node = q.front().first;
            int dist = q.front().second;
            if (!node->left && !node->right) {  // a leaf
                return node->val;
            }
            if (parent.count(node) > 0 && visited.count(parent[node]) == 0) {
                q.push(make_pair(parent[node], dist + 1));
                visited.insert(parent[node]);
            }
            if (node->left && visited.count(node->left) == 0) {
                q.push(make_pair(node->left, dist + 1));
                visited.insert(node->left);
            }
            if (node->right && visited.count(node->right) == 0) {
                q.push(make_pair(node->right, dist +1));
                visited.insert(node->right);
            }
            q.pop();
        }
        return -1;
    }
private:
    void constructParent(TreeNode* root, unordered_map<TreeNode*, TreeNode*> &parent, int k, TreeNode* &node) {
        if (root->val == k) {
            node = root;
        }
        if (root->left) {
            parent[root->left] = root;
            constructParent(root->left, parent, k, node);
        }
        if (root->right) {
            parent[root->right] = root;
            constructParent(root->right, parent, k, node);
        }
    }
};